Multiplicative minima
Let
and
be two
natural numbers
satisfying
.
Find the minimum possible value of
.
The answer is 658.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
2013 has the prime factorization 3 * 11 * 61 and the eight positive divisors: 1, 3, 11, 33, 61, 183, 671, 2013. We can write the original expression as a difference of squares:
2013 = b^2 - a^2 = (b+a)(b-a) (i).
and then calculate ordered pairs (a,b) according to the following systems of equations:
b + a = 2013; b - a = 1 => (a,b) = (1006, 1007); b + a = 671; b - a = 3 => (a,b) = (334, 337); b + a = 183; b - a = 11 => (a,b) = (86, 97); b + a = 61; b - a = 33 => (a,b) = (14, 47);
Of these four pairs, a = 14 & b = 47 yields the minimum product of 14 * 47 = 658.